DavidHolmes

• DavidHolmes 16 Aug 2019 8:11 UTC

  1 point

  in reply to: Stuart_Armstrong's comment on: Toy model piece #1: par­tial prefer­ences revisited

  (ac­tu­ally, my for­mula dou­bles the num­bers you gave)

  Are you sure? Sup­pose we take $W=W_1\bigsqcup W_2$ with $W_1=\{A,B,C,D\}$, $W_2=\{X,Y,Z\}$, then $n_1=3$, so the val­ues for $W_1$ should be $-3,-1,1,3$ as I gave them. And similarly for $W_2$, giv­ing val­ues $-2,0,2$. Or else I have mis-un­der­stood your defi­ni­tion?

  I’d sim­ply see that as two sep­a­rate par­tial preferences

  Just to be clear, by “sep­a­rate par­tial prefer­ence” you mean a sep­a­rate pre­order, on a set of ob­jects which may or may not have some over­lap with the ob­jects we con­sid­ered so far? Then some­how the work is just post­poned to the point where we try to com­bine par­tial prefer­ences?

  EDIT (in re­ply to your edit): I guess e.g. keep­ing con­di­tions 1,2,3 the same and in­stead min­imis­ing
  $g\left(G\right)={\sum }_{w\leftarrow w^{\prime }}{\lambda }_{w\leftarrow w^{\prime }}\left(U\right(w^{\prime })-U(w){)}^2,$
  where ${\lambda }_{w\leftarrow w^{\prime }}\in R_{>0}$ is pro­por­tion to the re­cip­ro­cal of the strength of the prefer­ence? Of course there are lots of var­i­ants on this!

• DavidHolmes 11 Aug 2019 16:38 UTC

  1 point

  on: Toy model piece #1: par­tial prefer­ences revisited

  This seems re­ally neat, but it seems quite sen­si­tive to how one defines the wor­lds un­der con­sid­er­a­tion, and whether one counts slightly differ­ent wor­lds as ac­tu­ally dis­tinct. Let me try to illus­trate this with an ex­am­ple.

  Sup­pose we have a $W$ con­sist­ing of 7 wor­lds, $W=\{A,B,C,D,X,Y,W\}$, with prefer­ences
  $$A<B<C<D,X<Y<Z$$ and no other non-triv­ial prefer­ences. Then (from the `sen­si­ble case’), I think we get the fol­low­ing util­ities:
  $A\mapsto -3$
  $X\mapsto -2$
  $B\mapsto -1$
  $Y\mapsto 0$
  $C\mapsto 1$
  $Z\mapsto 2$
  $D\mapsto 3$.

  Sup­pose now that I cre­ate two new copies $X^{\prime }$, $X^{\prime \prime }$ of the world $X$ which each differ by the po­si­tion of a sin­gle atom, so as to give me (ex­tremely weak!) prefer­ences $X^{\prime \prime }<X^{\prime }<X$, so all the non-triv­ial prefer­ences in the new $W$ are now sum­marised as $$A<B<C<D,X^{\prime \prime }<X^{\prime }<X<Y<Z.$$

  Then the re­sult­ing util­ities are (I think):
  $X^{\prime \prime }\mapsto -4$
  $A\mapsto -3$
  $X^{\prime }\mapsto -2$
  $B\mapsto -1$
  $X\mapsto 0$
  $C\mapsto 1$
  $Y\mapsto 2$
  $D\mapsto 3$
  $Z\mapsto 4$.

  In par­tic­u­lar, be­fore adding in these ‘triv­ial copies’ we had $U\left(Z\right)<U\left(D\right)$, and now we get $U\left(D\right)<U\left(Z\right)$. Is this a prob­lem? It de­pends on the situ­a­tion, but to me it sug­gests that, if us­ing this ap­proach, one needs to be care­ful in how the wor­lds are speci­fied, and the ‘fine-grained­ness’ needs to be roughly the same ev­ery­where.

• DavidHolmes 11 Aug 2019 16:31 UTC

  1 point

  in reply to: Stuart_Armstrong's comment on: Cat­e­go­rial prefer­ences and util­ity functions

  Thanks! I like the way your op­ti­mi­sa­tion prob­lem han­dles non-closed cy­cles.

  I think I’m less com­fortable with how it treats dis­con­nected com­po­nents—as I un­der­stand it you just trans­late each sep­a­rately to have `cen­tre of mass’ at 0. If one wants to get a util­ity func­tion out at the end one has to make some kind of choice in this situ­a­tion, and the choice you make is prob­a­bly the best one, so in that sense it seems very good.

  But for ex­am­ple it seems vuln­er­a­ble to cre­at­ing ‘vir­tual copies’ of wor­lds in or­der to shift the cen­tre of mass and push con­nected com­po­nents one way or the other. That was what started me think­ing about in­clud­ing strength of prefer­ence—if one adds to your setup a bunch of vir­tual copies of a world be­tween which one is `al­most in­differ­ent’ then it seems it will shift the cen­tre of mass, and thus the util­ity rel­a­tive to come other chain. Of course, if one is ac­tu­ally in­differ­ent then the ‘vir­tual copies’ will be col­lapsed to a sin­gle point in your $\overline{W}$, but if they are just ex­tremely close then it seems it will af­fect the util­ity rel­a­tive to some other chain. I’ll try to ex­plain this more clearly in a com­ment to your post.

• DavidHolmes 11 Aug 2019 15:59 UTC

  1 point

  in reply to: Charlie Steiner's comment on: Cat­e­go­rial prefer­ences and util­ity functions

  Thanks for the com­ment Char­lie.

  If I am in­differ­ent to a gam­ble with a prob­a­bil­ity $1$ of ice cream, and a prob­a­bil­ity 0.8 of choco­late cake and 0.2 of go­ing hungry

  To check I un­der­stand cor­rectly, you mean the agent is in­differ­ent be­tween the gam­bles (prob­a­bil­ity $1$ of ice cream) and (prob­a­bil­ity 0.8 of choco­late cake, prob­a­bil­ity 0.2 of go­ing hun­gry)?

  If I un­der­stand cor­rectly, you’re de­scribing a var­i­ant of Von Neu­mann–Mor­gen­stern where in­stead of giv­ing prefer­ences among all lot­ter­ies, you’re spec­i­fy­ing a cer­tain col­lec­tion of spe­cial type of pairs of lot­ter­ies be­tween which the agent is in­differ­ent${}^1$, to­gether with a sign to say in which `di­rec­tion’ things be­come preferred? It seems then likely to me that the data you give can be used to re­con­struct prefer­ences be­tween all lot­ter­ies...

  If one is given in­for­ma­tion in the form you pro­pose but only for an in­com­plete’ set of spe­cial triples (c.f.weak prefer­ences’ above), then one can again ask whether and in how many ways it can be ex­tended to a com­plete set of prefer­ences. It feels to me as if there is an ex­tra am­bi­guity com­ing in with your de­scrip­tion, for ex­am­ple if the set of pos­si­ble out­comes has 6 el­e­ments and I am given the value of the Bet­ter­ness func­tion on two dis­joint triples, then to gen­er­ate a util­ity func­tion I have to not only choose a `trans­la­tion’ be­tween the two triples, but also a scal­ing. But maybe this is bet­ter/​more re­al­is­tic!

  $1$. By `spe­cial types’, I mean in­differ­ence be­tween pairs of gam­bles of the form
  (prob­a­bil­ity $1$ of A) vs (prob­a­bil­ity $p$ of B and prob­a­bil­ity $(1-p)$ of C)
  for some $0\le p\le 1$, and pos­si­ble out­comes A, B, C. Then the sign says that I pre­fer higher prob­a­bil­ity of B (say).

• DavidHolmes 10 Aug 2019 0:16 UTC

  3 points

  on: Toy model piece #1: par­tial prefer­ences revisited

  Thanks for point­ing me to this up­dated ver­sion :-). This seems a re­ally neat trick for writ­ing down a util­ity func­tion that is com­pat­i­ble with the given pre­order. I thought a bit more about when/​to what ex­tent such a util­ity func­tion will be unique, in par­tic­u­lar if you are given not only the data of a pre­order, but also some in­for­ma­tion on the strengths of the prefer­ences. This ended up a bit too long for a com­ment, so I wrote a few things in out­line here:

  https://​​www.less­wrong.com/​​posts/​​7ncFy84ReMFW7TDG6/​​cat­e­go­rial-prefer­ences-and-util­ity-functions

  It may be quite ir­rele­vant to what you’re aiming for here, but I thought it was maybe worth writ­ing down just in case.

Cat­e­go­rial prefer­ences and util­ity functions

• DavidHolmes 9 Aug 2019 19:48 UTC

  1 point

  in reply to: Stuart_Armstrong's comment on: Par­tial prefer­ences and models

  Never mind—I had fun think­ing about this :-).

• DavidHolmes 9 Aug 2019 3:48 UTC

  12 points

  on: Par­tial prefer­ences and models

  Hi Stu­art,

  I’m work­ing my way through your `Re­search Agenda v0.9’ post, and am there­fore go­ing through var­i­ous older posts to un­der­stand things. I won­der if I could ask some ques­tions about the defi­ni­tion you pro­pose here?

  First, that $X$ be con­tained in $R^N$ for some $N$ seems not so rele­vant; can I just as­sume X, Y and Z are some man­i­folds ($C^k$ for some $0\le k\le \infty$)? And we are given some par­tial or­der $\prec$ on X, so that we can re­fer to `be­ing a bet­ter world’?

  Then, as I un­der­stand it, your defi­ni­tion says the fol­low­ing:

  Fix X, $\prec$ and Z. Let Y be a man­i­fold and $y_{+}$, $y_{-}\in Y$. Given a lo­cal ho­mo­mor­phism $+:Y\times Z\to X$, we say that $y_{+}$ is par­tially preferred to $y_{-}$ if for all $z\in Z$, we have $y_{-}+z\prec y_{+}+z$.

  I’m not sure which in­equal­ities should be strict, but this seems non-es­sen­tial for now. On the other hand, the de­pen­dence of this defi­ni­tion on the choice of Y seems some­what sub­tle and in­ter­est­ing. I will try to illus­trate this in what fol­lows.

  First, let us make a new defi­ni­tion. Fix X, $\prec$, and Z as be­fore. Let $Y^{\prime }=\{y_{+},y_{-}\}$, a two-el­e­ment set equipped with the dis­crete topol­ogy, and let ${+}^{\prime }:Y\times Z\to X$ be an im­mer­sion of $C^k$-man­i­folds. We say that $y_{+}$ is weakly par­tially preferred to $y_{-}$ if for all $z\in Z$, we have $y_{-}{+}^{\prime }z\prec y_{+}{+}^{\prime }z$.

  First, it is clear that par­tial prefer­ence im­plies weak par­tial prefer­ence. More for­mally:

  Claim 1: Fix X, $\prec$ and Z. Sup­pose we have a man­i­fold Y, points $y_{+}$, $y_{-}\in Y$, and a lo­cal ho­mo­mor­phism $+:Y\times Z\to X$ such that $y_{+}$ is par­tially preferred to $y_{-}$. Set­ting $Y^{\prime }=\{y_{+},y_{-}\}$ with the sub­space topol­ogy from $Y$ (i.e. dis­crete), and tak­ing ${+}^{\prime }$ to be the re­stric­tion of $+$ from $Y\times Z$ to $Y^{\prime }\times Z$, we have that $y_{+}$ is weakly par­tially preferred to $y_{-}$.

  Proof: ob­vi­ous. $\qed$

  How­ever, the con­verse can fail if Z is not con­tractible. First, let’s prove that the con­cepts are equiv­a­lent for Z con­tractible:

  Claim 2: Fix X, $\prec$ and Z, and as­sume that Z is con­tractible. Sup­pose we have a two-el­e­ment set $Y^{\prime }=\{y_{+},y_{-}\}$ and a map ${+}^{\prime }:Y^{\prime }\times Z\to X$ mak­ing $y_{+}$ weakly par­tially preferred to $y_{-}$. Then there ex­ist a man­i­fold Y, an in­jec­tion $Y^{\prime }\to Y$, and a lo­cal home­o­mor­phism $+:Y\times Z\to X$ whose re­stric­tion to $Y^{\prime }\times Z$ is ${+}^{\prime }$, mak­ing $y_{+}$ par­tially preferred to $y_{-}$.

  Proof: Let’s as­sume for sim­plic­ity of no­ta­tion that X is equidi­men­sional, say of di­men­sion $d_X$, and write $d_Z$ for the di­men­sion of Z. Let Y be the dis­joint union of two open balls of di­men­sion $d_X-d_Z$, with $Y^{\prime }\to Y$ the in­clu­sion of the cen­tres of the balls. Then take an $ϵ$-neigh­bour­hood of Z in X; it is diffeo­mor­phic to $Y\times Z$ since the nor­mal bun­dle to Z in X is triv­ial­is­able (c.f. https://​​math.stack­ex­change.com/​​ques­tions/​​857784/​​product-neigh­bor­hood-the­o­rem-with-bound­ary). $\qed$

  If we want ex­am­ples where weak par­tial prefer­ence and par­tial prefer­ence don’t co­in­cide, we should look for an ex­am­ple where Z is not con­tractible, and its nor­mal bun­dle in X is not con­tractible.

  Ex­am­ple 3: Let X be the dis­joint union of two moe­bius bands, and let Z be a cir­cle. Note that in­clud­ing Z along the cen­tre of ei­ther band gives a sub­man­i­fold whose tubu­lar neigh­bour­hood is not a product. As­sume that $\prec$ is such that one com­po­nent of X is preferred to the other (and $\prec$ is in­differ­ent within each con­nected com­po­nent). Then take $Y^{\prime }=\{y_{+},y_{-}\}$, and ${+}^{\prime }:Y^{\prime }\times Z\to X$ to be the in­clu­sion of the two cir­cles along the cen­tres of the two moe­bius bands, such that $\left\{y_{+}\right\}\times Z$ ends up in the preferred band. This yields a situ­a­tion where $y_{+}$ is weakly par­tially preferred to $y_{-}$, but the con­clu­sion of Claim 2 fails, i.e. this can­not be ex­tended to a par­tial prefer­ence for $y_{+}$ over $y_{-}$.

  What con­clu­sion should we draw from this? To me, it sug­gests that the no­tion of par­tial prefer­ence is not yet quite as one would want. In the set­ting of Ex­am­ple 3, where X con­sists of two moe­bius strips, one of which is preferred to the other, then land­ing in the preferred strip should be preferred to land­ing in the un-preferred strip?! And yet the `lo­cal home­o­mor­phism from a product’ con­di­tion gets in the way. This ex­am­ple is ob­vi­ously quite ar­tifi­cial, and maybe analo­gous things can­not oc­cur in re­al­ity. But I’m not so happy with this as an an­swer, since our ap­proaches to AI safety should be (so far as pos­si­ble) ro­bust against the flaws in our un­der­stand­ing of physics.

  Apolo­gies for the overly-long com­ment, and for the im­perfect LaTeX (I’ve not used this type of form much be­fore).

• DavidHolmes 17 Jun 2019 16:00 UTC

  3 points

  in reply to: Zack_M_Davis's comment on: The Uni­vari­ate Fallacy

  Thanks for the re­ply, Zack.

  The rea­son this ob­jec­tion doesn’t make the post com­pletely use­less...

  Sorry, I hope I didn’t sug­gest I thought that! You make a good point about some vari­ables be­ing more nat­u­ral in given ap­pli­ca­tions. I think it’s good to keep in mind that some­times it’s just a mat­ter of co­or­di­nate choice, and other times the points may be sep­a­rated but not in a lin­ear way.

• DavidHolmes 17 Jun 2019 2:14 UTC

  9 points

  on: The Uni­vari­ate Fallacy

  Hi Zack,

  Can you clar­ify some­thing? In the pic­ture you draw, there is a codi­men­sion-1 lin­ear sub­space sep­a­rat­ing the pa­ram­e­ter space into two halves, with all red points to one side, and all blue points to the other. Pro­ject­ing onto any 1-di­men­sional sub­space or­thog­o­nal to this (there is a unique one through the ori­gin) will thus yield a `vari­able’ which cleanly sep­a­rates the two points into the red and blue cat­e­gories. So in the illus­trated ex­am­ple, it looks just like a prob­lem of bad co­or­di­nate choice.

  On the other hand, one can eas­ily have much more patholog­i­cal situ­a­tions; for ex­am­ples, the red points could all lie in­side a cer­tain sphere, and the blue points out­side it. Then no choice of lin­ear co­or­di­nates will illus­trate this, and one has to use more ad­vanced anal­y­sis tech­niques to pick up on it (e.g. per­sis­tent ho­mol­ogy).

  So, to my vague ques­tion: do you have only the first situ­a­tion in mind, or are you also con­sid­er­ing the gen­eral case, but made the illus­trated ex­am­ple ex­tra-sim­ple?

  Per­haps this is clar­ified by your nu­mer­i­cal ex­am­ple, I’m afraid I’ve not checked.